Showing the divergence of $ \int_0^{\infty} \frac{1}{1+\sqrt{t}\sin(t)^2} dt$

Question

How can I show the divergence of

$$ \int_0^x \frac{1}{1+\sqrt{t}\sin(t)^2} dt$$

as $x\rightarrow\infty?$

Answer 1

For $t \gt 0$:

$$ 1 + t \ge 1 + \sqrt{t}\sin^2t $$

Or:

$$ \frac{1}{1 + t} \le \frac{1}{1 + \sqrt{t}\sin^2t} $$

Now consider:

$$ \int_0^x \frac{dt}{1 + t} \le \int_0^x \frac{dt}{1 + \sqrt{t}\sin^2t} $$

The LHS diverges as $x \to +\infty$, so the RHS does too.

Answer 2

Use $1+\sqrt{t}\sin^2(t)\leqslant1+\sqrt{t}$ uniformly over $t$.